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Forces B/w Dislocations Consider two parallel (//) edge dislocations lying in the same slip plane. The two dislocations can be of same sign or different signs (a) Same Sign (on same slip plane)

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When the two dislocations are separated by large distance: The total elastic energy per unit length of the dislocation is given by: When dislocations are very close together: The arrangement can be considered approximately a single dislocation with Burgers vector = 2b In order to reduce the total elastic energy, same sign dislocations will repel each other (i.e., prefer large distance separation). (14.34) (14.35)

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Dislocations of opposite sign (on same slip plane) If the dislocations are separated by large distance: If dislocations are close together: Burgers vector = b - b = 0 Hence, in order to reduce their total energy, dislocations of opposite signs will prefer to come together and annihilate (cancel) each other.

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The same conclusions are obtained for dislocation of mixed orientations (a) and (b) above can be summarized as: –Like dislocations repel and –unlike dislocations attract

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Dislocations Not on the Same Slip Plane Consider two dislocations lying parallel to the z (x 3 ) -axis: In order to solve this: (a) We assume that dislocation “I” is at the origin (b) We then find the interaction force on dislocation “II” due to dislocation “I” x2x2 x1x1 x3x3 I II

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Recall Eqn Note that the dislocation at the origin (dislocation I) provides the stress field, while the Burgers vector and the dislocation length belongs to dislocation II Since is edge: Also b II is parallel to x 1 : Therefore, This means that b 2 = b 3 = 0 and b 1 = b 14.29

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Since t II is parallel to x 3, then This means that t 1 = t 2 = 0 and t 3 = 1 From Eqn , we can write: Therefore

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But Therefore, F along the x 1 Direction is given as: 21 b This component of force is responsible for dislocation glide motion - i.e., for dislocation II to move along x 1 axis

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F along the x 2 Direction is given as: - 11 b This component of force is responsible for climb (along x 2 ). At ambient (low) temperature, F x2 is not important (because, no climb). For edge dislocation, movement is by slip & slip occurs only in the plane contained by the dislocation line & its Burgers vector

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Consider only component F x1 For x 1 >0: F x1 is negative (attractive) when x 1

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Hence Stable positions for two edge dislocations

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Equations and can also be obtained by considering both the radial and tangential components. The force per unit length is given by: Because edge dislocations are mainly confined to the plane, the force component along the the x direction, which is the slip direction, is of most interest, and is given by:

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Eqn is same as Figure 14-5 is a plot of the variation of F x with distance x, using equation Where x is expressed in units of y. Curve A is for dislocations of the same sign; curve B is for dislocations of opposite sign. Note that dislocations of the same sign repel each other when x > y, and attract each other when x < y

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Figure Graphical representation of Eq. (14-21). Solid curve A is for two edge dislocations of same sign. Dashed curve B is for two unlike two dislocations.

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Example: A dislocations lies parallel to [100] with Burgers vector b. Compute the force acting on the dislocation due to the stress field of a neighboring screw dislocation lying parallel to [001]. Assume that for the screw dislocations x2x2 x1x1 x3x3 Solution: S

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Let the screw dislocation be dislocation I at the origin. The stress field for screw dislocation is given by: based on the assumption, we have

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For the other dislocation

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(continued)

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(b) (a) Figure (a) Diffusion of vacancy to edge dislocation; (b) dislocation climbs up one lattice spacing

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